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Theorem sspsstri 4033
Description: Two ways of stating trichotomy with respect to inclusion. (Contributed by NM, 12-Aug-2004.)
Assertion
Ref Expression
sspsstri ((𝐴𝐵𝐵𝐴) ↔ (𝐴𝐵𝐴 = 𝐵𝐵𝐴))

Proof of Theorem sspsstri
StepHypRef Expression
1 or32 923 . 2 (((𝐴𝐵𝐵𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ 𝐵𝐴))
2 sspss 4030 . . . 4 (𝐴𝐵 ↔ (𝐴𝐵𝐴 = 𝐵))
3 sspss 4030 . . . . 5 (𝐵𝐴 ↔ (𝐵𝐴𝐵 = 𝐴))
4 eqcom 2808 . . . . . 6 (𝐵 = 𝐴𝐴 = 𝐵)
54orbi2i 910 . . . . 5 ((𝐵𝐴𝐵 = 𝐴) ↔ (𝐵𝐴𝐴 = 𝐵))
63, 5bitri 278 . . . 4 (𝐵𝐴 ↔ (𝐵𝐴𝐴 = 𝐵))
72, 6orbi12i 912 . . 3 ((𝐴𝐵𝐵𝐴) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ (𝐵𝐴𝐴 = 𝐵)))
8 orordir 927 . . 3 (((𝐴𝐵𝐵𝐴) ∨ 𝐴 = 𝐵) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ (𝐵𝐴𝐴 = 𝐵)))
97, 8bitr4i 281 . 2 ((𝐴𝐵𝐵𝐴) ↔ ((𝐴𝐵𝐵𝐴) ∨ 𝐴 = 𝐵))
10 df-3or 1085 . 2 ((𝐴𝐵𝐴 = 𝐵𝐵𝐴) ↔ ((𝐴𝐵𝐴 = 𝐵) ∨ 𝐵𝐴))
111, 9, 103bitr4i 306 1 ((𝐴𝐵𝐵𝐴) ↔ (𝐴𝐵𝐴 = 𝐵𝐵𝐴))
Colors of variables: wff setvar class
Syntax hints:  wb 209  wo 844  w3o 1083   = wceq 1538  wss 3884  wpss 3885
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2114  ax-9 2122  ax-ext 2773
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-3or 1085  df-ex 1782  df-sb 2070  df-clab 2780  df-cleq 2794  df-clel 2873  df-ne 2991  df-v 3446  df-in 3891  df-ss 3901  df-pss 3903
This theorem is referenced by:  ordtri3or  6195  sorpss  7438  sorpssi  7439  funpsstri  33116
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